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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 241-270 of 555 questions

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241 Medium
241 of 555
Application +4 | -1
The angle between the lines represented by the equation (x2+y2)sin⁡θ+2xy=0\left(x^2+y^2\right) \sin \theta+2 x y=0 is
View answer and explanation
Correct option: C
π2−θ\frac{\pi}{2}-\theta
Option C (π2−θ\frac{\pi}{2}-\theta) is correct because it is calculated by using the formula for the angle between the lines and trigonometric identities.
242 Easy
242 of 555
Application +4 | -1
The orthocentre of the triangle formed by the lines x=2,y=3x=2, y=3 and 3x+2y=63 x+2 y=6 is at the point
View answer and explanation
Correct option: C
(2,3)(2,3)
Option C (2,3) is correct because the orthocentre of a right triangle is the vertex where the right angle is formed, which is the intersection of x=2 and y=3.
243 Easy
243 of 555
Analyzing +4 | -1
The distance between the two points AA and A′A^{\prime} which lie on y=2y=2 such that both the line segments ABA B and A′BA^{\prime} B (where BB is the point (2,3)(2,3) ) subtend angle π4\frac{\pi}{4} at the origin, is equal to
View answer and explanation
Correct option: C
525\frac{52}{5}
The correct answer is B (525\frac{{52}}{{5}}) because it represents the distance between points A and A' calculated using the tangent of the angle and the given point B.
244 Easy
244 of 555
Application +4 | -1
Two lines are drawn through (3,4)(3,4), each of which makes angle of 45∘45^{\circ} with the line x−y=2x-y=2, then area of the triangle formed by these lines is NN. Find 2N2 N.
View answer and explanation
Correct option: A
9
The correct answer is A (9) because it is the value of 2N after finding the area of the triangle formed by the given lines.
245 Easy
245 of 555
Analyzing +4 | -1
If the straight lines x+3y=4,3x+y=4x+3 y=4,3 x+y=4 and x+y=0x+y=0 form a triangle, then the triangle is
View answer and explanation
Correct option: D
right angled isosceles
Option D (right angled isosceles) is correct because the triangle formed by the lines has two equal sides and a right angle.
246 Easy
246 of 555
Application +4 | -1
Let A(1,1),B(6,5),C(3/2,2)A(1,1), B(6,5), C(3 / 2,2) be the vertices of △ABC\triangle A B C. If a point PP is such that areas of △PAB,△PAC\triangle P A B, \triangle P A C and △PBC\triangle P B C are equal and Q(−76,−13)Q\left(-\frac{7}{6},-\frac{1}{3}\right) is a point, then PQ=P Q=
View answer and explanation
Correct option: D
5
Option D (5) is correct because point P is the centroid of the triangle, and the distance PQ is calculated using the distance formula.
247 Easy
247 of 555
Application +4 | -1
If a point P(4,  3)P\left( {4,\;3} \right) is shifted by a distance2\sqrt 2 unit parallel to the line y=xy = x, then coordinates of p in new position are
View answer and explanation
Correct option: A
(5, 4)
Option D (30.36) is correct because the strength of 10 volume H2O2 solution is calculated to be approximately 30.36 g/L.
248 Easy
248 of 555
Analyzing +4 | -1
The equation of the sides of a triangle are x−3y=0x - 3y = 0, 4x+3y=54x + 3y = 5 and 3x+y=03x + y = 0. The line 3x−4y=03x - 4y = 0 passes through
View answer and explanation
Correct option: C
The orthocentre
Option C (The orthocentre) is correct because the intersection of the perpendicular lines x - 3y = 0 and 3x + y = 0 is the orthocenter (0,0), and the line 3x - 4y = 0 passes through it.
249 Easy
249 of 555
Analyzing +4 | -1
The straight line x+2y−9=0,3x+5y−5=0x+2 y-9=0,3 x+5 y-5=0 and ax+by−1=0a x+b y-1=0 are concurrent if the straight line 35x−22y+1=035 x-22 y+1=0 passes through the point
View answer and explanation
Correct option: A
(a,b)(a, b)
Option A ((a,b)(a, b)) is correct because the concurrency condition leads to an equation that the line 35x−22y+1=035x - 22y + 1 = 0 satisfies at the point (a,b)(a, b).
250 Medium
250 of 555
Application +4 | -1
If OO is the origin and QQ is a variable point on x2=4yx^2=4 y, then the locus of the mid-point of OQO Q, is
View answer and explanation
Correct option: D
x2=2yx^2=2 y
Option D is correct. If OO is the origin and QQ is a variable point on x2=4yx^2=4y, then the locus of the mid-point of OQOQ is x2=2yx^2=2y.
251 Easy
251 of 555
Application +4 | -1
If the point (a,a) falls between the lines ∣x+y∣=4\left| {x + y} \right| = 4, then
View answer and explanation
Correct option: A
∣a∣=2\left| a \right| = 2
Option C (∣a∣<2|a| < 2) is correct because the point (a, a) must lie between the lines x + y = 4 and x + y = -4.
252 Medium
252 of 555
Application +4 | -1
The line which is parallel to x-axis and crosses the curve y=x  y = \sqrt x \; at an angle 45∘45^\circ, is
View answer and explanation
Correct option: B
y=12y = \frac{1}{2}
Option B (y=12y = \frac{1}{2}) is correct because it represents a line parallel to the x-axis that intersects the curve at a 45-degree angle.
253 Easy
253 of 555
Application +4 | -1
The circumcentre of the triangle with vertices (0, 30), (4, 0) and (30, 0) is
View answer and explanation
Correct option: D
(17, 17)
Option D (17, 17) is correct because the circumcenter is equidistant from all vertices of the triangle, and solving the equations based on this condition yields the coordinates (17, 17).
254 Easy
254 of 555
Application +4 | -1
The equation of a straight line parallel to 2x+3y+11=02 x+3 y+11=0 and which is such that the sum of its intercept on the axes is 15 , is
View answer and explanation
Correct option: A
2x+3y−18=02 x+3 y-18=0
Option A (2x+3y−18=02 x+3 y-18=0) is correct because it represents a line parallel to the given line with intercepts summing to 15.
255 Easy
255 of 555
Understanding +4 | -1
Consider the following statements about straight lines:

I. Slope of horizontal line is zero and slope of vertical line is undefined.

II. Two lines are parallel if and only if their slopes are equal.

III. Two lines are perpendicular if and only if product of their slope is −1.-1 .

Which of the above statements are true?
View answer and explanation
Correct option: D
All the above
Option D is correct because all three statements about straight lines are fundamental and well-established geometric principles.
256 Easy
256 of 555
Application +4 | -1
Let the base of a triangle lie along the line x=a and be of length 2a. The area of this triangle is a2{a^2} if the vertex lies on the lines
View answer and explanation
Correct option: B
x=0,  x=ax = 0,\;x = a
Option A (x=−a,  x=2ax = - a,\;x = 2a) is incorrect. The correct answer should be x=0,  x=2ax = 0,\;x = 2a because these lines satisfy the area condition.
257 Easy
257 of 555
Analyzing +4 | -1
The locus of a point represented by x=a2(t+1t),y=a2(t−1t)  x = \frac{a}{2}\left( {\frac{{t + 1}}{t}} \right),y = \frac{a}{2}\left( {\frac{{t - 1}}{t}} \right)\; is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the locus of the point is a straight line, which is not among the given options.
258 Easy
258 of 555
Analyzing +4 | -1
The equation of a straight line passing through (1,2) and having intercept of length 3 between the straight lines 3x+4y=243x + 4y = 24 and 3x  4y=123x\;4y = 12 is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because after finding the equation of the line passing through (1,2) with a length of 3 intercepted between the given lines, it does not match any of the provided options.
259 Easy
259 of 555
Application +4 | -1
The line x+y=4x + y = 4 divides the line joining (−1,1)\left( { - 1,1} \right) and (5,7) in the ratio λ:1\lambda :1, then the value of λ\lambda is
View answer and explanation
Correct option: B
01-Feb
Option B (0.5) is correct because it is the ratio in which the line divides the line segment joining the two points, calculated using the section formula.
260 Medium
260 of 555
Application +4 | -1
A line having slope &#39;1&#39; is drawn from a point A(−3,0)A(-3,0) cuts a curve y=x2+x+1y=x^2+x+1 at P & QP \,\&\, Q. Find ℓ(AP)ℓ(AQ)\ell({AP}) \ell({AQ}).
View answer and explanation
Correct option: A
14
Option A (14) is correct because it is the product of the distances from the point A to the intersection points P and Q, found by solving the quadratic equation and applying Vieta's formulas.
261 Medium
261 of 555
Application +4 | -1
The line x−2y=0x-2 y=0 will be a bisector of the angle between the lines represented by the equation x2−2hxy−2y2=0x^2-2 h x y-2 y^2=0, if h=h=
View answer and explanation
Correct option: C
−2-2
Option C (−2-2) is correct because the line x - 2y = 0 is a bisector, and using the properties of angle bisectors and the given equation, we can solve for h.
262 Easy
262 of 555
Analyzing +4 | -1
If a variable line drawn through the intersection of the lines x3+y4=1\frac{x}{3}+\frac{y}{4}=1 and x4+y3=1\frac{x}{4}+\frac{y}{3}=1, meets the coordinates axes at AA and BB, then the locus of the mid-point of ABA B, is
View answer and explanation
Correct option: A
7xy=6(x+y)7 x y=6(x+y)
Option A (7xy=6(x+y)7 x y=6(x+y)) is correct because it is the locus of the midpoint of AB, derived from the intersection of the given lines.
263 Easy
263 of 555
Application +4 | -1
Points (1,2)(1,2) and (−2,1)(-2,1) are
View answer and explanation
Correct option: C
On the opposite side of 4x+2y=14 x+2 y=1
Option C (On the opposite side of 4x+2y=14 x+2 y=1) is correct because substituting the points into the equation 4x+2y−14x+2y-1 yields opposite signs.
264 Easy
264 of 555
Application +4 | -1
A line passing through the point P(1,2)\mathrm{P}(1,2) meets the line x+y=7x+y=7 at the distance of 3 units from PP. Then the slope of this line satisfies the equation:
View answer and explanation
Correct option: B
7x2−18x+7=07 x^2-18 x+7=0
Option B (7x2−18x+7=07 x^2-18 x+7=0) is correct because the slope of the line satisfies this quadratic equation, derived from the given conditions.
265 Medium
265 of 555
Application +4 | -1
If the line px−qy=rpx - qy = r intersects the coordinate axes at (a,0) and (0,b), thyen the value of a+b is equal to
View answer and explanation
Correct option: B
r(q−ppq)r\left( {\frac{{q - p}}{{pq}}} \right)
Option A (r(q+ppq)r\left( {\frac{{{{q + p}}}}{{{{pq}}}}} \right)) is correct because it represents the sum of the x and y intercepts of the line px - qy = r.
266 Medium
266 of 555
Application +4 | -1
A line through the point A(2,4)A(2,4) intersects the line x+y=9x+y=9 at the point P.P. The minimum distance of AP,A P, is
View answer and explanation
Correct option: C
32\frac{3}{\sqrt{2}}
Option C (32\frac{3}{\sqrt{2}}) is correct because it represents the minimum distance AP, which is the perpendicular distance from A to the line x+y=9.
267 Easy
267 of 555
Application +4 | -1
If the lines x+3y−9=0,  4x+by−2=0x + 3y - 9 = 0,\;4x + by - 2 = 0 and 2x−y−4=02x - y - 4 = 0 are concurrent, then b equals
View answer and explanation
Correct option: A
-5
Option A (-5) is correct because the determinant of the coefficients must be zero for concurrent lines.
268 Easy
268 of 555
Application +4 | -1
The straight lines ax+by=c,  bx+cy=aax + by = c,\;bx + cy = a and cx+ay=bcx + ay = b are concurrent, if
View answer and explanation
Correct option: D
a+b+c=0a + b + c = 0
Option D (a+b+c=0a + b + c = 0) is correct because it satisfies the condition for concurrency of three lines, derived from setting the determinant of the coefficients to zero.
269 Easy
269 of 555
Application +4 | -1
The area enclosed by the lines ∣x∣+∣y∣=2|x|+|y|=2 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the area enclosed by |x| + |y| = 2 is 8 square units, which is not among the given options.
270 Easy
270 of 555
Application +4 | -1
The slopes of the lines which make an angle 45∘{45^ \circ } with the line 3x−y=−53x - y = - 5 are
View answer and explanation
Correct option: D
−2,  12- 2,\;\frac{1}{2}
Option A (x+2y−3=0x + 2y - 3 = 0) is correct because it represents a line parallel to the given lines and divides the distance between them in the specified ratio.
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